Given Eigenvectors and Eigenvalues, Compute a Matrix Product (Stanford University Exam)

Stanford University Linear Algebra Exam Problems and Solutions

Problem 181

Suppose that $\begin{bmatrix}
1 \\
1
\end{bmatrix}$ is an eigenvector of a matrix $A$ corresponding to the eigenvalue $3$ and that $\begin{bmatrix}
2 \\
1
\end{bmatrix}$ is an eigenvector of $A$ corresponding to the eigenvalue $-2$.
Compute $A^2\begin{bmatrix}
4 \\
3
\end{bmatrix}$.

(Stanford University Linear Algebra Exam Problem)
 
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Solution.

By the given conditions, we have
\[A\begin{bmatrix}
1 \\
1
\end{bmatrix}=3\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } A \begin{bmatrix}
2 \\
1
\end{bmatrix}=-2\begin{bmatrix}
2 \\
1
\end{bmatrix}. \tag{*}\]

We express the vector $\begin{bmatrix}
4 \\
3
\end{bmatrix}$ as a linear combination of vectors $\begin{bmatrix}
1 \\
1
\end{bmatrix}$ and $\begin{bmatrix}
2 \\
1
\end{bmatrix}$. Let
\[a\begin{bmatrix}
1 \\
1
\end{bmatrix}+b\begin{bmatrix}
2 \\
1
\end{bmatrix}
=\begin{bmatrix}
4 \\
3
\end{bmatrix}\] and we want to determine scalars $a$ and $b$. The augmented matrix of the system is reduced as follows:
\begin{align*}
\left[\begin{array}{rr|r}
1 & 2 & 4 \\
1 &1 &3
\end{array}\right] \to
\left[\begin{array}{rr|r}
1 & 0& 2 \\
0 &1 & 1
\end{array}\right].
\end{align*}

Thus the solution is
\[a=2, b=1\] and hence we obtain the linear combination
\[2\begin{bmatrix}
1 \\
1
\end{bmatrix}+\begin{bmatrix}
2 \\
1
\end{bmatrix}
=\begin{bmatrix}
4 \\
3
\end{bmatrix}.\]


Before computing $A^2\begin{bmatrix}
4 \\
3
\end{bmatrix}$, we first calculate $A\begin{bmatrix}
4 \\
3
\end{bmatrix}$ as follows.
\begin{align*}
A\begin{bmatrix}
4 \\
3
\end{bmatrix}&= A \left(2\begin{bmatrix}
1 \\
1
\end{bmatrix}+\begin{bmatrix}
2 \\
1
\end{bmatrix} \right)\\
&=2A\begin{bmatrix}
1 \\
1
\end{bmatrix}+A\begin{bmatrix}
2 \\
1
\end{bmatrix}\\
&=2\cdot 3\begin{bmatrix}
1 \\
1
\end{bmatrix}+(-2)\begin{bmatrix}
2 \\
1
\end{bmatrix} \text{ by } (*)\\
&=6\begin{bmatrix}
1 \\
1
\end{bmatrix}-2\begin{bmatrix}
2 \\
1
\end{bmatrix}.
\end{align*}


Now, using the result of the above computation, we compute
\begin{align*}
A^2 \begin{bmatrix}
4 \\
3
\end{bmatrix}&= A\left( \begin{bmatrix}
4 \\
3
\end{bmatrix}\right)\\
&=A\left( 6\begin{bmatrix}
1 \\
1
\end{bmatrix}-2\begin{bmatrix}
2 \\
1
\end{bmatrix} \right)\\
&=6A\begin{bmatrix}
1 \\
1
\end{bmatrix}-2A\begin{bmatrix}
2 \\
1
\end{bmatrix}\\
&=6 \cdot 3\begin{bmatrix}
1 \\
1
\end{bmatrix}-2(-2)\begin{bmatrix}
2 \\
1
\end{bmatrix} \text{ by } (*)\\
&=\begin{bmatrix}
26 \\
22
\end{bmatrix}.
\end{align*}

Therefore we obtained
\[A^2\begin{bmatrix}
4 \\
3
\end{bmatrix}=\begin{bmatrix}
26 \\
22
\end{bmatrix}.\]


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